When it reaches it's peak, the energy is converted into potential as it slows down, then back to kinetic as it goes back to the lowest point.
Based on your problem where as ask for the distance of the ball drop between the pitchers mound and the home plate and with a given of the speed of ball is 43m/s and the homeplates is 60.6ft away. Based on my step by step procedure and also considering the value of gravity by 9.8m/s^2 i came up with the distance of 144m away
Answer:
The maximum height the box will reach is 1.72 m
Explanation:
F = k·x
Where
F = Force of the spring
k = The spring constant = 300 N/m
x = Spring compression or stretch = 0.15 m
Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N
Mass of box = 0.2 kg
Work, W, done by the spring = 
and the kinetic energy gained by the box is given by KE = 
Since work done by the spring = kinetic energy gained by the box we have
= therefore we have v = 
= 
= 
= 5.81 m/s
Therefore the maximum height is given by
v² = 2·g·h or h = 
= 
= 1.72 m
Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.
Answer:
The final velocity of the ball is 7m/s
Explanation:
M1=8kg, V1 =10m/s
, M2=2kg
, V2=-5m/s
initial momentum before collison
m1v1+m2v2
=8×10 +2×(-5) =80-10 = 70kg m/s
final momentum after collison
=(m1+m2)×v
=(8+2)×v
=10v
According to the law of conversion of momentum
initial momentum =final momentum
70=10v
10v=70
v=70/10
v=7m/s
